Question: Divide the following complex numbers. $ \dfrac{9+13i}{-1+3i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-1-3i}$ $ \dfrac{9+13i}{-1+3i} = \dfrac{9+13i}{-1+3i} \cdot \dfrac{{-1-3i}}{{-1-3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(9+13i) \cdot (-1-3i)} {(-1+3i) \cdot (-1-3i)} = \dfrac{(9+13i) \cdot (-1-3i)} {(-1)^2 - (3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(9+13i) \cdot (-1-3i)} {(-1)^2 - (3i)^2} = $ $ \dfrac{(9+13i) \cdot (-1-3i)} {1 + 9} = $ $ \dfrac{(9+13i) \cdot (-1-3i)} {10} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({9+13i}) \cdot ({-1-3i})} {10} = $ $ \dfrac{{9} \cdot {(-1)} + {13} \cdot {(-1) i} + {9} \cdot {-3 i} + {13} \cdot {-3 i^2}} {10} $ Evaluate each product of two numbers. $ \dfrac{-9 - 13i - 27i - 39 i^2} {10} $ Finally, simplify the fraction. $ \dfrac{-9 - 13i - 27i + 39} {10} = \dfrac{30 - 40i} {10} = 3-4i $